Electricity and Magnetism

Transcription

1 Electricit nd Mgnetism prt 4 We now stud the brnch of phsics concerned with electric nd mgnetic phenomen. The lws of electricit nd mgnetism pl centrl role in the opertion of such devices s MP3 plers, televisions, electric motors, computers, high-energ ccelertors, nd other electronic devices. More fundmentll, the intertomic nd intermoleculr forces responsible for the formtion of solids nd liuids re electric in origin. Evidence in Chinese documents suggests mgnetism ws observed s erl s 2000 BC. The ncient Greeks observed electric nd mgnetic phenomen possibl s erl s 700 BC. The Greeks knew bout mgnetic forces from observtions tht the nturll occurring stone mgnetite (Fe 3 O 4 ) is ttrcted to iron. (The word electric comes from elecktron, the Greek A Trnsrpid mglev trin pulls into sttion in hnghi, word for mber. The word mgnetic comes from Mgnesi, the nme of the district of Greece where mgnetite ws Chin. The word mglev is n bbrevited form of mgnetic levittion. This trin mkes no phsicl contct with its rils; its first found.) weight is totll supported b electromgnetic forces. In this Not until the erl prt of the nineteenth centur did scientists estblish tht electricit nd mgnetism re relted prt of the book, we will stud these forces. (OTHK/Asi Imges/ Jupiterimges) phenomen. In 1819, Hns Oersted discovered tht compss needle is deflected when plced ner circuit crring n electric current. In 1831, Michel Frd nd, lmost simultneousl, Joseph Henr showed tht when wire is moved ner mgnet (or, euivlentl, when mgnet is moved ner wire), n electric current is estblished in the wire. In 1873, Jmes Clerk Mwell used these observtions nd other eperimentl fcts s bsis for formulting the lws of electromgnetism s we know them tod. (Electromgnetism is nme given to the combined stud of electricit nd mgnetism.) Mwell s contributions to the field of electromgnetism were especill significnt becuse the lws he formulted re bsic to ll forms of electromgnetic phenomen. His work is s importnt s Newton s work on the lws of motion nd the theor of grvittion. 657

2 chpter 23 Electric Fields 23.1 Properties of Electric Chrges 23.2 Chrging Objects b Induction 23.3 Coulomb s Lw 23.4 The Electric Field 23.5 Electric Field of Continuous Chrge Distribution 23.6 Electric Field Lines 23.7 Motion of Chrged Prticle in Uniform Electric Field In this chpter, we begin the stud of electromgnetism. The link to our previous stud is through the concept of force. The electromgnetic force between chrged prticles is one of the fundmentl forces of nture. We begin b describing some bsic properties of one mnifesttion of the electromgnetic force, the electric force. We then discuss Coulomb s lw, which is the fundmentl lw governing the electric force between n two chrged prticles. Net, we introduce the concept of Mother nd dughter re both enjoing the effects of electricll n electric field ssocited with chrge distribution nd chrging their bodies. Ech individul hir on their heds becomes describe its effect on other chrged prticles. We then chrged nd eerts repulsive force on the other hirs, resulting in the stnd-up hirdos seen here. (Courtes of Resonnce Reserch show how to use Coulomb s lw to clculte the electric Corportion) field for given chrge distribution. The chpter concludes with discussion of the motion of chrged prticle in uniform electric field Properties of Electric Chrges A number of simple eperiments demonstrte the eistence of electric forces. For emple, fter rubbing blloon on our hir on dr d, ou will find tht the blloon ttrcts bits of pper. The ttrctive force is often strong enough to suspend the pper from the blloon. When mterils behve in this w, the re sid to be electrified or to hve become electricll chrged. You cn esil electrif our bod b vigorousl rubbing our

3 23.1 Properties of Electric Chrges 659 A negtivel chrged rubber rod suspended b string is ttrcted to positivel chrged glss rod. A negtivel chrged rubber rod is repelled b nother negtivel chrged rubber rod. Figure 23.1 The electric force between () oppositel chrged objects nd (b) like-chrged objects. F F Glss Rubber b F Rubber F Rubber shoes on wool rug. Evidence of the electric chrge on our bod cn be detected b lightl touching (nd strtling) friend. Under the right conditions, ou will see sprk when ou touch nd both of ou will feel slight tingle. (Eperiments such s these work best on dr d becuse n ecessive mount of moisture in the ir cn cuse n chrge ou build up to lek from our bod to the Erth.) In series of simple eperiments, it ws found tht there re two kinds of electric chrges, which were given the nmes positive nd negtive b Benjmin Frnklin ( ). Electrons re identified s hving negtive chrge, nd protons re positivel chrged. To verif tht there re two tpes of chrge, suppose hrd rubber rod tht hs been rubbed on fur is suspended b string s shown in Figure When glss rod tht hs been rubbed on silk is brought ner the rubber rod, the two ttrct ech other (Fig. 23.1). On the other hnd, if two chrged rubber rods (or two chrged glss rods) re brought ner ech other s shown in Figure 23.1b, the two repel ech other. This observtion shows tht the rubber nd glss hve two different tpes of chrge on them. On the bsis of these observtions, we conclude tht chrges of the sme sign repel one nother nd chrges with opposite signs ttrct one nother. Using the convention suggested b Frnklin, the electric chrge on the glss rod is clled positive nd tht on the rubber rod is clled negtive. Therefore, n chrged object ttrcted to chrged rubber rod (or repelled b chrged glss rod) must hve positive chrge, nd n chrged object repelled b chrged rubber rod (or ttrcted to chrged glss rod) must hve negtive chrge. Another importnt spect of electricit tht rises from eperimentl observtions is tht electric chrge is lws conserved in n isolted sstem. Tht is, when one object is rubbed ginst nother, chrge is not creted in the process. The electrified stte is due to trnsfer of chrge from one object to the other. One object gins some mount of negtive chrge while the other gins n eul mount of positive chrge. For emple, when glss rod is rubbed on silk s in Figure 23.2, the silk obtins negtive chrge eul in mgnitude to the positive chrge on the glss rod. We now know from our understnding of tomic structure tht electrons re trnsferred in the rubbing process from the glss to the silk. imilrl, when rubber is rubbed on fur, electrons re trnsferred from the fur to the rubber, giving the rubber net negtive chrge nd the fur net positive chrge. This process works becuse neutrl, unchrged mtter contins s mn positive chrges (protons within tomic nuclei) s negtive chrges (electrons). In 1909, Robert Millikn ( ) discovered tht electric chrge lws occurs s integrl multiples of fundmentl mount of chrge e (see ection 25.7). In modern terms, the electric chrge is sid to be untized, where is the stndrd smbol used for chrge s vrible. Tht is, electric chrge eists s discrete Becuse of conservtion of chrge, ech electron dds negtive chrge to the silk nd n eul positive chrge is left on the glss rod. Figure 23.2 When glss rod is rubbed with silk, electrons re trnsferred from the glss to the silk. Also, becuse the chrges re trnsferred in discrete bundles, the chrges on the two objects re 6e, or 62e, or 63e, nd so on. Electric chrge is conserved

4 660 CHAPTER 23 Electric Fields The neutrl sphere hs eul numbers of positive nd negtive chrges. Electrons redistribute when chrged rod is brought close. b ome electrons leve the grounded sphere through the ground wire. The ecess positive chrge is nonuniforml distributed. The remining electrons redistribute uniforml, nd there is net uniform distribution of positive chrge on the sphere. c d Figure 23.3 Chrging metllic object b induction. () A neutrl metllic sphere. (b) A chrged rubber rod is plced ner the sphere. (c) The sphere is grounded. (d) The ground connection is removed. (e) The rod is removed. e pckets, nd we cn write 5 6Ne, where N is some integer. Other eperiments in the sme period showed tht the electron hs chrge 2e nd the proton hs chrge of eul mgnitude but opposite sign 1e. ome prticles, such s the neutron, hve no chrge. Quick Quiz 23.1 Three objects re brought close to ech other, two t time. When objects A nd B re brought together, the repel. When objects B nd C re brought together, the lso repel. Which of the following re true? () Objects A nd C possess chrges of the sme sign. (b) Objects A nd C possess chrges of opposite sign. (c) All three objects possess chrges of the sme sign. (d) One object is neutrl. (e) Additionl eperiments must be performed to determine the signs of the chrges Chrging Objects b Induction It is convenient to clssif mterils in terms of the bilit of electrons to move through the mteril: Electricl conductors re mterils in which some of the electrons re free electrons 1 tht re not bound to toms nd cn move reltivel freel through the mteril; electricl insultors re mterils in which ll electrons re bound to toms nd cnnot move freel through the mteril. Mterils such s glss, rubber, nd dr wood fll into the ctegor of electricl insultors. When such mterils re chrged b rubbing, onl the re rubbed becomes chrged nd the chrged prticles re unble to move to other regions of the mteril. In contrst, mterils such s copper, luminum, nd silver re good electricl conductors. When such mterils re chrged in some smll region, the chrge redil distributes itself over the entire surfce of the mteril. emiconductors re third clss of mterils, nd their electricl properties re somewhere between those of insultors nd those of conductors. ilicon nd germnium re well-known emples of semiconductors commonl used in the fbriction of vriet of electronic chips used in computers, cellulr telephones, nd home theter sstems. The electricl properties of semiconductors cn be chnged over mn orders of mgnitude b the ddition of controlled mounts of certin toms to the mterils. To understnd how to chrge conductor b process known s induction, consider neutrl (unchrged) conducting sphere insulted from the ground s shown in Figure There re n eul number of electrons nd protons in the sphere if the chrge on the sphere is ectl zero. When negtivel chrged rubber rod is brought ner the sphere, electrons in the region nerest the rod eperience repulsive force nd migrte to the opposite side of the sphere. This migrtion leves the side of the sphere ner the rod with n effective positive chrge becuse of the diminished number of electrons s in Figure 23.3b. (The left side of the sphere in Figure 23.3b is positivel chrged s if positive chrges moved into this region, but remember tht onl electrons re free to move.) This process occurs even if the rod never ctull touches the sphere. If the sme eperiment is performed with conducting wire connected from the sphere to the Erth (Fig. 23.3c), some of the electrons in the conductor re so strongl repelled b the presence of the negtive chrge in the rod tht the move out of the sphere through the wire nd into the Erth. The smbol t the end of the wire in Figure 23.3c indictes tht the wire 1 A metl tom contins one or more outer electrons, which re wekl bound to the nucleus. When mn toms combine to form metl, the free electrons re these outer electrons, which re not bound to n one tom. These electrons move bout the metl in mnner similr to tht of gs molecules moving in continer.

5 23.3 Coulomb's Lw 661 The chrged blloon induces chrge seprtion on the surfce of the wll due to relignment of chrges in the molecules of the wll. The chrged rod ttrcts the pper becuse chrge seprtion is induced in the molecules of the pper. Wll Induced chrge seprtion Chrged blloon b. Cengge Lerning/Chrles D. Winters Figure 23.4 () A chrged blloon is brought ner n insulting wll. (b) A chrged rod is brought close to bits of pper. is connected to ground, which mens reservoir, such s the Erth, tht cn ccept or provide electrons freel with negligible effect on its electricl chrcteristics. If the wire to ground is then removed (Fig. 23.3d), the conducting sphere contins n ecess of induced positive chrge becuse it hs fewer electrons thn it needs to cncel out the positive chrge of the protons. When the rubber rod is removed from the vicinit of the sphere (Fig. 23.3e), this induced positive chrge remins on the ungrounded sphere. Notice tht the rubber rod loses none of its negtive chrge during this process. Chrging n object b induction reuires no contct with the object inducing the chrge. Tht is in contrst to chrging n object b rubbing (tht is, b conduction), which does reuire contct between the two objects. A process similr to induction in conductors tkes plce in insultors. In most neutrl molecules, the center of positive chrge coincides with the center of negtive chrge. In the presence of chrged object, however, these centers inside ech molecule in n insultor m shift slightl, resulting in more positive chrge on one side of the molecule thn on the other. This relignment of chrge within individul molecules produces ler of chrge on the surfce of the insultor s shown in Figure The proimit of the positive chrges on the surfce of the object nd the negtive chrges on the surfce of the insultor results in n ttrctive force between the object nd the insultor. Your knowledge of induction in insultors should help ou eplin wh chrged rod ttrcts bits of electricll neutrl pper s shown in Figure 23.4b. Quick Quiz 23.2 Three objects re brought close to one nother, two t time. When objects A nd B re brought together, the ttrct. When objects B nd C re brought together, the repel. Which of the following re necessril true? () Objects A nd C possess chrges of the sme sign. (b) Objects A nd C possess chrges of opposite sign. (c) All three objects possess chrges of the sme sign. (d) One object is neutrl. (e) Additionl eperiments must be performed to determine informtion bout the chrges on the objects Coulomb s Lw Chrles Coulomb mesured the mgnitudes of the electric forces between chrged objects using the torsion blnce, which he invented (Fig. 23.5). The operting principle of the torsion blnce is the sme s tht of the pprtus used b Cvendish to mesure the grvittionl constnt (see ection 13.1), with the electricll neutrl B A uspension hed Fiber Figure 23.5 Coulomb s torsion blnce, used to estblish the inversesure lw for the electric force between two chrges.

6 662 CHAPTER 23 Electric Fields spheres replced b chrged ones. The electric force between chrged spheres A nd B in Figure 23.5 cuses the spheres to either ttrct or repel ech other, nd the resulting motion cuses the suspended fiber to twist. Becuse the restoring torue of the twisted fiber is proportionl to the ngle through which the fiber rottes, mesurement of this ngle provides untittive mesure of the electric force of ttrction or repulsion. Once the spheres re chrged b rubbing, the electric force between them is ver lrge compred with the grvittionl ttrction, nd so the grvittionl force cn be neglected. From Coulomb s eperiments, we cn generlize the properties of the electric force (sometimes clled the electrosttic force) between two sttionr chrged prticles. We use the term point chrge to refer to chrged prticle of zero size. The electricl behvior of electrons nd protons is ver well described b modeling them s point chrges. From eperimentl observtions, we find tht the mgnitude of the electric force (sometimes clled the Coulomb force) between two point chrges is given b Coulomb s lw. AIP Emilio egre Visul Archives, E. cott Brr Collection Coulomb s lw Coulomb constnt Chrles Coulomb French phsicist ( ) Coulomb s mjor contributions to science were in the res of electrosttics nd mgnetism. During his lifetime, he lso investigted the strengths of mterils nd determined the forces tht ffect objects on bems, thereb contributing to the field of structurl mechnics. In the field of ergonomics, his reserch provided fundmentl understnding of the ws in which people nd nimls cn best do work. F e 5 k e (23.1) r 2 where k e is constnt clled the Coulomb constnt. In his eperiments, Coulomb ws ble to show tht the vlue of the eponent of r ws 2 to within n uncertint of few percent. Modern eperiments hve shown tht the eponent is 2 to within n uncertint of few prts in Eperiments lso show tht the electric force, like the grvittionl force, is conservtive. The vlue of the Coulomb constnt depends on the choice of units. The I unit of chrge is the coulomb (C). The Coulomb constnt k e in I units hs the vlue This constnt is lso written in the form k e N? m 2 /C 2 (23.2) k e 5 1 (23.3) 4pP 0 where the constnt P 0 (Greek letter epsilon) is known s the permittivit of free spce nd hs the vlue P C 2 /N? m 2 (23.4) The smllest unit of free chrge e known in nture, 2 the chrge on n electron (2e) or proton (1e), hs mgnitude e C (23.5) Therefore, 1 C of chrge is pproimtel eul to the chrge of electrons or protons. This number is ver smll when compred with the number of free electrons in 1 cm 3 of copper, which is on the order of Nevertheless, 1 C is substntil mount of chrge. In tpicl eperiments in which rubber or glss rod is chrged b friction, net chrge on the order of C is obtined. In other words, onl ver smll frction of the totl vilble chrge is trnsferred between the rod nd the rubbing mteril. The chrges nd msses of the electron, proton, nd neutron re given in Tble Notice tht the electron nd proton re identicl in the mgnitude of their chrge but vstl different in mss. On the other hnd, the proton nd neutron re similr in mss but vstl different in chrge. Chpter 46 will help us understnd these interesting properties. 2 No unit of chrge smller thn e hs been detected on free prticle; current theories, however, propose the eistence of prticles clled urks hving chrges 2e/3 nd 2e/3. Although there is considerble eperimentl evidence for such prticles inside nucler mtter, free urks hve never been detected. We discuss other properties of urks in Chpter 46.

7 23.3 Coulomb's Lw 663 TABLE 23.1 Chrge nd Mss of the Electron, Proton, nd Neutron Prticle Chrge (C) Mss (kg) Electron (e) Proton (p) Neutron (n) Emple 23.1 The Hdrogen Atom The electron nd proton of hdrogen tom re seprted (on the verge) b distnce of pproimtel m. Find the mgnitudes of the electric force nd the grvittionl force between the two prticles. OLUTION Conceptulize Think bout the two prticles seprted b the ver smll distnce given in the problem sttement. In Chpter 13, we found the grvittionl force between smll objects to be wek, so we epect the grvittionl force between the electron nd proton to be significntl smller thn the electric force. Ctegorize The electric nd grvittionl forces will be evluted from universl force lws, so we ctegorize this emple s substitution problem. Use Coulomb s lw to find the mgnitude of the electric force: F e 5 k 0 e 002e0 e N? m 2 /C C2 2 r m N Use Newton s lw of universl grvittion nd Tble 23.1 (for the prticle msses) to find the mgnitude of the grvittionl force: F g 5 G m em p r N? m 2 /kg kg kg m N The rtio F e /F g < Therefore, the grvittionl force between chrged tomic prticles is negligible when compred with the electric force. Notice the similr forms of Newton s lw of universl grvittion nd Coulomb s lw of electric forces. Other thn the mgnitude of the forces between elementr prticles, wht is fundmentl difference between the two forces? When deling with Coulomb s lw, remember tht force is vector untit nd must be treted ccordingl. Coulomb s lw epressed in vector form for the electric force eerted b chrge 1 on second chrge 2, written F 12, is 12 F 5 k 1 2 e r 2 r^ 12 (23.6) where r^ 12 is unit vector directed from 1 towrd 2 s shown in Active Figure Becuse the electric force obes Newton s third lw, the electric force eerted Vector form of Coulomb s lw When the chrges re of the sme sign, the force is repulsive. When the chrges re of opposite signs, the force is ttrctive. F 21 1 r rˆ 12 2 F 12 b 1 F 21 F 12 2 ACTIVE FIGURE 23.6 Two point chrges seprted b distnce r eert force on ech other tht is given b Coulomb s lw. The force F 21 eerted b 2 on 1 is eul in mgnitude nd opposite in direction to the force F 12 eerted b 1 on 2.

8 664 CHAPTER 23 Electric Fields b 2 on 1 is eul in mgnitude to the force eerted b 1 on 2 nd in the opposite direction; tht is, F 21 52F 12. Finll, Eution 23.6 shows tht if 1 nd 2 hve the sme sign s in Active Figure 23.6, the product 1 2 is positive nd the electric force on one prticle is directed w from the other prticle. If 1 nd 2 re of opposite sign s shown in Active Figure 23.6b, the product 1 2 is negtive nd the electric force on one prticle is directed towrd the other prticle. These signs describe the reltive direction of the force but not the bsolute direction. A negtive product indictes n ttrctive force, nd positive product indictes repulsive force. The bsolute direction of the force on chrge depends on the loction of the other chrge. For emple, if n is lies long the two chrges in Active Figure 23.6, the product 1 2 is positive, but F 12 points in the positive direction nd F 21 points in the negtive direction. When more thn two chrges re present, the force between n pir of them is given b Eution Therefore, the resultnt force on n one of them euls the vector sum of the forces eerted b the other individul chrges. For emple, if four chrges re present, the resultnt force eerted b prticles 2, 3, nd 4 on prticle 1 is F1 5 F21 1 F31 1 F41 Quick Quiz 23.3 Object A hs chrge of 12 mc, nd object B hs chrge of 16 mc. Which sttement is true bout the electric forces on the objects? () FAB 523F BA (b) FAB 52F BA (c) 3F AB 52F BA (d) FAB 5 3F BA (e) FAB 5 FBA (f) 3F AB 5 FBA Emple 23.2 Find the Resultnt Force Consider three point chrges locted t the corners of right tringle s shown in Figure 23.7, where mc, mc, nd m. Find the resultnt force eerted on 3. OLUTION 2 F 23 3 F 13 Conceptulize Think bout the net force on 3. Becuse chrge 3 is ner two other chrges, it will eperience two electric forces. These forces re eerted in different directions s shown in Figure Ctegorize Becuse two forces re eerted on chrge 3, we ctegorize this emple s vector ddition problem. 1 Anlze The directions of the individul forces eerted b 1 nd 2 on 3 re shown in Figure The force F 23 eerted b 2 on 3 is ttrctive becuse 2 nd 3 hve opposite signs. In the coordinte sstem shown in Figure 23.7, the ttrctive force F 23 is to the left (in the negtive direction). The force F 13 eerted b 1 on 3 is repulsive becuse both chrges re positive. The repulsive force F 13 mkes n ngle of with the is. Figure 23.7 (Emple 23.2) The force eerted b 1 on 3 is F 13. The force eerted b 2 on 3 is F 23. The resultnt force F 3 eerted on 3 is the vector sum F 13 1 F 23. Use Eution 23.1 to find the mgnitude of F23: F 23 5 k e Find the mgnitude of the force F13: F 13 5 k e 1 " N? m 2 /C C C m N N? m 2 /C C C m N

9 23.3 Coulomb's Lw cont. Find the nd components of the force F 13: Find the components of the resultnt force cting on 3 : Epress the resultnt force cting on 3 in unit-vector form: F 13 5 F 13 cos N F 13 5 F 13 sin N F 3 5 F 13 1 F N 1 (28.99 N) N F 3 5 F 13 1 F N N F i^ j^ 2 N Finlize The net force on 3 is upwrd nd towrd the left in Figure If 3 moves in response to the net force, the distnces between 3 nd the other chrges chnge, so the net force chnges. Therefore, if 3 is free to move, it cn be modeled s prticle under net force s long s it is recognized tht the force eerted on 3 is not constnt. As reminder, we displ most numericl vlues to three significnt figures, which leds to opertions such s 7.94 N 1 (28.99 N) N bove. If ou crr ll intermedite results to more significnt figures, ou will see tht this opertion is correct. WHAT IF? for F3? Wht if the signs of ll three chrges were chnged to the opposite signs? How would tht ffect the result Answer The chrge 3 would still be ttrcted towrd 2 nd repelled from 1 with forces of the sme mgnitude. Therefore, the finl result for F 3 would be the sme. Emple 23.3 Where Is the Net Force Zero? Three point chrges lie long the is s shown in Figure The positive chrge mc is t m, the positive chrge mc is t the origin, nd the net force cting on 3 is zero. Wht is the coordinte of 3? OLUTION Conceptulize Becuse 3 is ner two other chrges, it eperiences two electric forces. Unlike Figure 23.8 (Emple 23.3) Three point chrges re plced long the is. If the resultnt force cting on 3 is zero, the force F13 eerted b 1 on 3 must be eul in mgnitude nd opposite in direction to the force F23 eerted b 2 on 3. the preceding emple, however, the forces lie long the sme line in this problem s indicted in Figure Becuse 3 is negtive nd 1 nd 2 re positive, the forces F 13 nd F 23 re both ttrctive. Ctegorize Becuse the net force on 3 is zero, we model the point chrge s prticle in euilibrium. 2 F m F 13 1 Anlze Write n epression for the net force on chrge 3 when it is in euilibrium: Move the second term to the right side of the eution nd set the coefficients of the unit vector i^ eul: Eliminte k e nd u 3 u nd rerrnge the eution: 3 F 5 F23 1 F13 52k e 2 k e k e i^ 1 k e i^ 5 0 ( ) 2 u 2 u 5 2 u 1 u ( )( C) 5 2 ( C) Reduce the udrtic eution to simpler form: olve the udrtic eution for the positive root: m Finlize The second root to the udrtic eution is m. Tht is nother loction where the mgnitudes of the forces on 3 re eul, but both forces re in the sme direction. continued

10 666 CHAPTER 23 Electric Fields 23.3 cont. WHAT IF? uppose 3 is constrined to move onl long the is. From its initil position t m, it is pulled smll distnce long the is. When relesed, does it return to euilibrium, or is it pulled frther from euilibrium? Tht is, is the euilibrium stble or unstble? Answer If 3 is moved to the right, F 13 becomes lrger nd F 23 becomes smller. The result is net force to the right, in the sme direction s the displcement. Therefore, the chrge 3 would continue to move to the right nd the euilibrium is unstble. (ee ection 7.9 for review of stble nd unstble euilibri.) If 3 is constrined to st t fied coordinte but llowed to move up nd down in Figure 23.8, the euilibrium is stble. In this cse, if the chrge is pulled upwrd (or downwrd) nd relesed, it moves bck towrd the euilibrium position nd oscilltes bout this point. Emple 23.4 Find the Chrge on the pheres Two identicl smll chrged spheres, ech hving mss of kg, hng in euilibrium s shown in Figure The length L of ech string is m, nd the ngle u is Find the mgnitude of the chrge on ech sphere. OLUTION Conceptulize Figure 23.9 helps us conceptulize this emple. The two spheres eert repulsive forces on ech other. If the re held close to ech other nd relesed, the move outwrd from the center nd settle into the configurtion in Figure 23.9 fter the oscilltions hve vnished due to ir resistnce. Ctegorize The ke phrse in euilibrium helps us model ech sphere s prticle in euilibrium. This emple is similr to the prticle in euilibrium problems in Chpter 5 with the dded feture tht one of the forces on sphere is n electric force. L u u L F e T cos u b u mg T u T sin u Figure 23.9 (Emple 23.4) () Two identicl spheres, ech crring the sme chrge, suspended in euilibrium. (b) Digrm of the forces cting on the sphere on the left prt of (). Anlze The force digrm for the left-hnd sphere is shown in Figure 23.9b. The sphere is in euilibrium under the ppliction of the force T from the string, the electric force F e from the other sphere, nd the grvittionl force m g. Write Newton s second lw for the left-hnd sphere in component form: (1) o F 5 T sin u 2 F e 5 0 T sin u 5 F e (2) o F 5 T cos u 2 mg 5 0 T cos u 5 mg Divide Eution (1) b Eution (2) to find F e : Use the geometr of the right tringle in Figure 23.9 to find reltionship between, L, nd u: tn u 5 F e mg sin u 5 L F e 5 mg tn u 5 L sin u olve Coulomb s lw (E. 23.1) for the chrge uu on ech sphere: F 0 0 e r 2 F e mg tn u 12L sin u Å k e Å k e Å k e kg m/s 2 2 tn m2 sin ubstitute numericl vlues: Å N? m 2 /C C

11 23.4 The Electric Field cont. Finlize If the sign of the chrges were not given in Figure 23.9, we could not determine them. In fct, the sign of the chrge is not importnt. The sitution is the sme whether both spheres re positivel chrged or negtivel chrged. WHAT IF? uppose our roommte proposes solving this problem without the ssumption tht the chrges re of eul mgnitude. he clims the smmetr of the problem is destroed if the chrges re not eul, so the strings would mke two different ngles with the verticl nd the problem would be much more complicted. How would ou respond? Answer The smmetr is not destroed nd the ngles re not different. Newton s third lw reuires the mgnitudes of the electric forces on the two spheres to be the sme, regrdless of the eulit or noneulit of the chrges. The solution to the emple remins the sme with one chnge: the vlue of uu in the solution is replced b " in the new sitution, where 1 nd 2 re the vlues of the chrges on the two spheres. The smmetr of the problem would be destroed if the msses of the spheres were not the sme. In this cse, the strings would mke different ngles with the verticl nd the problem would be more complicted The Electric Field In ection 5.1, we discussed the differences between contct forces nd field forces. Two field forces the grvittionl force in Chpter 13 nd the electric force here hve been introduced into our discussions so fr. As pointed out erlier, field forces cn ct through spce, producing n effect even when no phsicl contct occurs between intercting objects. The grvittionl field g t point in spce due to source prticle ws defined in ection 13.4 to be eul to the grvittionl force Fg cting on test prticle of mss m divided b tht mss: g ; Fg /m. The concept of field ws developed b Michel Frd ( ) in the contet of electric forces nd is of such prcticl vlue tht we shll devote much ttention to it in the net severl chpters. In this pproch, n electric field is sid to eist in the region of spce round chrged object, the source chrge. When nother chrged object the test chrge enters this electric field, n electric force cts on it. As n emple, consider Figure 23.10, which shows smll positive test chrge 0 plced ner second object crring much greter positive chrge Q. We define the electric field due to the source chrge t the loction of the test chrge to be the electric force on the test chrge per unit chrge, or, to be more specific, the electric field vector E t point in spce is defined s the electric force Fe cting on positive test chrge 0 plced t tht point divided b the test chrge: 3 Fe E ; (23.7) 0 The vector E hs the I units of newtons per coulomb (N/C). The direction of E s shown in Figure is the direction of the force positive test chrge eperiences when plced in the field. Note tht E is the field produced b some chrge or chrge distribution seprte from the test chrge; it is not the field produced b the test chrge itself. Also note tht the eistence of n electric field is propert of its source; the presence of the test chrge is not necessr for the field to eist. The test chrge serves s detector of the electric field: n electric field eists t point if test chrge t tht point eperiences n electric force. Eution 23.7 cn be rerrnged s Fe 5 E (23.8) 3 When using Eution 23.7, we must ssume the test chrge 0 is smll enough tht it does not disturb the chrge distribution responsible for the electric field. If the test chrge is gret enough, the chrge on the metllic sphere is redistributed nd the electric field it sets up is different from the field it sets up in the presence of the much smller test chrge. This drmtic photogrph cptures lightning bolt striking tree ner some rurl homes. Lightning is ssocited with ver strong electric fields in the tmosphere. Definition of electric field Q 0 P E Test chrge ource chrge Figure A smll positive test chrge 0 plced t point P ner n object crring much lrger positive chrge Q eperiences n electric field E t point P estblished b the source chrge Q. We will lws ssume tht the test chrge is so smll tht the field of the source chrge is unffected b its presence. Courtes Johnn Auter

12 668 CHAPTER 23 Electric Fields Pitfll Prevention 23.1 Prticles Onl Eution 23.8 is vlid onl for prticle of chrge, tht is, n object of zero size. For chrged object of finite size in n electric field, the field m vr in mgnitude nd direction over the size of the object, so the corresponding force eution m be more complicted. This eution gives us the force on chrged prticle plced in n electric field. If is positive, the force is in the sme direction s the field. If is negtive, the force nd the field re in opposite directions. Notice the similrit between Eution 23.8 nd the corresponding eution for prticle with mss plced in grvittionl field, F g 5 m g (ection 5.5). Once the mgnitude nd direction of the electric field re known t some point, the electric force eerted on n chrged prticle plced t tht point cn be clculted from Eution To determine the direction of n electric field, consider point chrge s source chrge. This chrge cretes n electric field t ll points in spce surrounding it. A test chrge 0 is plced t point P, distnce r from the source chrge, s in Active Figure We imgine using the test chrge to determine the direction of the electric force nd therefore tht of the electric field. According to Coulomb s lw, the force eerted b on the test chrge is e F 5 k 0 e r 2 r^ where r^ is unit vector directed from towrd 0. This force in Active Figure is directed w from the source chrge. Becuse the electric field t P, the position of the test chrge, is defined b E 5 Fe / 0, the electric field t P creted b is E 5 k e r 2 r^ (23.9) If the source chrge is positive, Active Figure 23.11b shows the sitution with the test chrge removed: the source chrge sets up n electric field t P, directed w from. If is negtive s in Active Figure 23.11c, the force on the test chrge is towrd the source chrge, so the electric field t P is directed towrd the source chrge s in Active Figure 23.11d. To clculte the electric field t point P due to group of point chrges, we first clculte the electric field vectors t P individull using Eution 23.9 nd then dd them vectorill. In other words, t n point P, the totl electric field due to group of source chrges euls the vector sum of the electric fields of ll the chrges. This superposition principle pplied to fields follows directl from the vector ddition of electric forces. Therefore, the electric field t point P due to group of source chrges cn be epressed s the vector sum Electric field due to finite number of point chrges i E 5 k e i 2 r r^ i (23.10) i where r i is the distnce from the ith source chrge i to the point P nd r^ i is unit vector directed from i towrd P. If is positive, the force on the test chrge 0 is directed w from. rˆ r 0 P F e rˆ F e 0 P If is negtive, the force on the test chrge 0 is directed towrd. c ACTIVE FIGURE (), (c) When test chrge 0 is plced ner source chrge, the test chrge eperiences force. (b), (d) At point P ner source chrge, there eists n electric field. For positive source chrge, the electric field t P points rdill outwrd from. b rˆ P E d rˆ E P For negtive source chrge, the electric field t P points rdill inwrd towrd.

13 23.4 The Electric Field 669 In Emple 23.5, we eplore the electric field due to two chrges using the superposition principle. Prt (B) of the emple focuses on n electric dipole, which is defined s positive chrge nd negtive chrge 2 seprted b distnce 2. The electric dipole is good model of mn molecules, such s hdrochloric cid (HCl). Neutrl toms nd molecules behve s dipoles when plced in n eternl electric field. Furthermore, mn molecules, such s HCl, re permnent dipoles. The effect of such dipoles on the behvior of mterils subjected to electric fields is discussed in Chpter 26. Quick Quiz 23.4 A test chrge of 13 mc is t point P where n eternl electric field is directed to the right nd hs mgnitude of N/C. If the test chrge is replced with nother test chrge of 23 mc, wht hppens to the eternl electric field t P? () It is unffected. (b) It reverses direction. (c) It chnges in w tht cnnot be determined. Emple 23.5 Electric Field Due to Two Chrges Chrges 1 nd 2 re locted on the is, t distnces nd b, respectivel, from the origin s shown in Figure E 1 (A) Find the components of the net electric field t the point P, which is t position (0, ). f E OLUTION Conceptulize Compre this emple with Emple There, we dd vector forces to find the net force on chrged prticle. Here, we dd electric field vectors to find the net electric field t point in spce. Ctegorize We hve two source chrges nd wish to find the resultnt electric field, so we ctegorize this emple s one in which we cn use the superposition principle represented b Eution Figure (Emple 23.5) The totl electric field E t P euls the vector sum E1 1 E2, where E1 is the field due to the positive chrge 1 nd E2 is the field due to the negtive chrge 2. P r 1 f 1 u E 2 b u r 2 2 Anlze Find the mgnitude of the electric field t P due to chrge 1 : E 1 5 k e r 5 k 2 e Find the mgnitude of the electric field t P due to chrge 2 : E 2 5 k e r 5 k 2 e b Write the electric field vectors for ech chrge in unit-vector form: Write the components of the net electric field vector: E1 5 k e E2 5 k e cos f i^ 1 k 2 e b 2 1 cos u i^ 2 k 2 e sin f j^ sin u j^ b (1) E 5 E 1 1 E 2 5 k e 2 1 cos f1k e b 2 1 cos u 2 (2) E 5 E 1 1 E 2 5 k e sin f2k e b 2 1 sin u 2 continued

14 670 CHAPTER 23 Electric Fields 23.5 cont. (B) Evlute the electric field t point P in the specil cse tht u 1 u 5 u 2 u nd 5 b. E 1 OLUTION Conceptulize Figure shows the sitution in this specil cse. Notice the smmetr in the sitution nd tht the chrge distribution is now n electric dipole. Ctegorize Becuse Figure is specil cse of the generl cse shown in Figure 23.12, we cn ctegorize this emple s one in which we cn tke the result of prt (A) nd substitute the pproprite vlues of the vribles. Figure (Emple 23.5) When the chrges in Figure re of eul mgnitude nd euidistnt from the origin, the sitution becomes smmetric s shown here. P r u u u E 2 E u Anlze Bsed on the smmetr in Figure 23.13, evlute Eutions (1) nd (2) from prt (A) with 5 b, u 1 u 5 u 2 u 5, nd f 5 u: From the geometr in Figure 23.13, evlute cos u: ubstitute Eution (4) into Eution (3): (3) E 5 k e 2 1 cos u1k 2 e 2 1 cos u52k 2 e 2 1 cos u 2 E 5 k e 2 1 sin u2k 2 e sin u (4) cos u 5 r /2 E 5 2k e /2 5 k e /2 (C) Find the electric field due to the electric dipole when point P is distnce.. from the origin. OLUTION In the solution to prt (B), becuse.., neglect 2 compred with 2 nd write the epression for E in this cse: (5) E < k e 2 3 Finlize From Eution (5), we see tht t points fr from dipole but long the perpendiculr bisector of the line joining the two chrges, the mgnitude of the electric field creted b the dipole vries s 1/r 3, wheres the more slowl vring field of point chrge vries s 1/r 2 (see E. 23.9). Tht is becuse t distnt points, the fields of the two chrges of eul mgnitude nd opposite sign lmost cncel ech other. The 1/r 3 vrition in E for the dipole lso is obtined for distnt point long the is nd for n generl distnt point Electric Field of Continuous Chrge Distribution Ver often, the distnces between chrges in group of chrges re much smller thn the distnce from the group to point where the electric field is to be clculted. In such situtions, the sstem of chrges cn be modeled s continuous. Tht is, the sstem of closel spced chrges is euivlent to totl chrge tht is continuousl distributed long some line, over some surfce, or throughout some volume.

15 23.5 Electric Field of Continuous Chrge Distribution 671 To set up the process for evluting the electric field creted b continuous chrge distribution, let s use the following procedure. First, divide the chrge distribution into smll elements, ech of which contins smll chrge D s shown in Figure Net, use Eution 23.9 to clculte the electric field due to one of these elements t point P. Finll, evlute the totl electric field t P due to the chrge distribution b summing the contributions of ll the chrge elements (tht is, b ppling the superposition principle). The electric field t P due to one chrge element crring chrge D is DE 5 k e D r 2 r^ where r is the distnce from the chrge element to point P nd r^ is unit vector directed from the element towrd P. The totl electric field t P due to ll elements in the chrge distribution is pproimtel E < k e i D i r i 2 r^ i where the inde i refers to the ith element in the distribution. Becuse the chrge distribution is modeled s continuous, the totl field t P in the limit D i 0 is ˆr 2 r 2 2 r 1 P E E 3 2 E 1 rˆ 1 r3 ˆr Figure The electric field t P due to continuous chrge distribution is the vector sum of the fields DE i due to ll the elements D i of the chrge distribution. Three smple elements re shown. E 5 k e lim D i 0 i D i 2 r r^ i 5 k d e 3 i r 2 r^ (23.11) where the integrtion is over the entire chrge distribution. The integrtion in Eution is vector opertion nd must be treted ppropritel. Let s illustrte this tpe of clcultion with severl emples in which the chrge is distributed on line, on surfce, or throughout volume. When performing such clcultions, it is convenient to use the concept of chrge densit long with the following nottions: If chrge Q is uniforml distributed throughout volume V, the volume chrge densit r is defined b r ; Q V where r hs units of coulombs per cubic meter (C/m 3 ). If chrge Q is uniforml distributed on surfce of re A, the surfce chrge densit s (Greek letter sigm) is defined b s ; Q A where s hs units of coulombs per sure meter (C/m 2 ). If chrge Q is uniforml distributed long line of length,, the liner chrge densit l is defined b Electric field due to continuous chrge distribution Volume chrge densit urfce chrge densit l ; Q, where l hs units of coulombs per meter (C/m). If the chrge is nonuniforml distributed over volume, surfce, or line, the mounts of chrge d in smll volume, surfce, or length element re d 5 r dv d 5 s da d 5 l d, Liner chrge densit

16 672 CHAPTER 23 Electric Fields CALCULATING THE ELECTRIC FIELD Problem-olving trteg The following procedure is recommended for solving problems tht involve the determintion of n electric field due to individul chrges or chrge distribution. 1. Conceptulize. Estblish mentl representtion of the problem: think crefull bout the individul chrges or the chrge distribution nd imgine wht tpe of electric field it would crete. Appel to n smmetr in the rrngement of chrges to help ou visulize the electric field. 2. Ctegorize. Are ou nlzing group of individul chrges or continuous chrge distribution? The nswer to this uestion tells ou how to proceed in the Anlze step. 3. Anlze. () If ou re nlzing group of individul chrges, use the superposition principle: when severl point chrges re present, the resultnt field t point in spce is the vector sum of the individul fields due to the individul chrges (E ). Be ver creful in the mnipultion of vector untities. It m be useful to review the mteril on vector ddition in Chpter 3. Emple 23.5 demonstrted this procedure. (b) If ou re nlzing continuous chrge distribution, replce the vector sums for evluting the totl electric field from individul chrges b vector integrls. The chrge distribution is divided into infinitesiml pieces, nd the vector sum is crried out b integrting over the entire chrge distribution (E ). Emples 23.6 through 23.8 demonstrte such procedures. Consider smmetr when deling with either distribution of point chrges or continuous chrge distribution. Tke dvntge of n smmetr in the sstem ou observed in the Conceptulize step to simplif our clcultions. The cncelltion of field components perpendiculr to the is in Emple 23.7 is n emple of the ppliction of smmetr. 4. Finlize. Check to see if our electric field epression is consistent with the mentl representtion nd if it reflects n smmetr tht ou noted previousl. Imgine vring prmeters such s the distnce of the observtion point from the chrges or the rdius of n circulr objects to see if the mthemticl result chnges in resonble w. Emple 23.6 The Electric Field Due to Chrged Rod A rod of length, hs uniform positive chrge per unit length l nd totl chrge Q. Clculte the electric field t point P tht is locted long the long is of the rod nd distnce from one end (Fig ). OLUTION Conceptulize The field de t P due to ech segment of chrge on the rod is in the negtive direction becuse ever segment crries positive chrge. Ctegorize Becuse the rod is continuous, we re evluting the field due to continuous chrge distribution rther thn group of individul chrges. Becuse ever segment of the rod produces n electric field in the negtive direction, the sum of their contributions cn be hndled without the need to dd vectors. Anlze Let s ssume the rod is ling long the is, d is the length of one smll segment, nd d is the chrge on tht segment. Becuse the rod hs chrge per unit length l, the chrge d on the smll segment is d 5 l d. E P d Figure (Emple 23.6) The electric field t P due to uniforml chrged rod ling long the is.

17 23.5 Electric Field of Continuous Chrge Distribution cont. Find the mgnitude of the electric field t P due to one segment of the rod hving chrge d: Find the totl field t P using 4 Eution 23.11: E 5 3,1 de 5 k d e 5 k l d 2 e 2 k e l d 2 Noting tht k e nd l 5 Q /, re constnts nd cn be removed from the integrl, evlute the integrl: E 5 k e l 3,1 d 2 5 k e l c2 1 d,1 (1) E 5 k e Q, 1 2 1, 1 b 5 k e Q 1, 1 2 Finlize If 0, which corresponds to sliding the br to the left until its left end is t the origin, then E `. Tht represents the condition in which the observtion point P is t zero distnce from the chrge t the end of the rod, so the field becomes infinite. WHAT IF? uppose point P is ver fr w from the rod. Wht is the nture of the electric field t such point? Answer If P is fr from the rod (..,), then, in the denomintor of Eution (1) cn be neglected nd E < k e Q/ 2. Tht is ectl the form ou would epect for point chrge. Therefore, t lrge vlues of /,, the chrge distribution ppers to be point chrge of mgnitude Q ; the point P is so fr w from the rod we cnnot distinguish tht it hs size. The use of the limiting techniue (/, `) is often good method for checking mthemticl epression. Emple 23.7 The Electric Field of Uniform Ring of Chrge A ring of rdius crries uniforml distributed positive totl chrge Q. Clculte the electric field due to the ring t point P ling distnce from its center long the centrl is perpendiculr to the plne of the ring (Fig ). OLUTION Conceptulize Figure shows the electric field contribution de t P due to single segment of chrge t the top of the ring. This field vector cn be resolved into components de prllel to the is of the ring nd de perpendiculr to the is. Figure 23.16b shows the electric field contributions from two d u P de segments on opposite sides of the ring. Becuse of the smmetr of the sitution, the perpendiculr components of the field cncel. Tht is true for ll pirs of segments round the ring, so we cn ignore the perpendiculr component of the field nd focus solel on the prllel components, which simpl dd. Ctegorize Becuse the ring is continuous, we re evluting the field due to continuous chrge distribution rther thn group of individul chrges. continued r de de b 2 1 de 2 u Figure (Emple 23.7) A uniforml chrged ring of rdius. () The field t P on the is due to n element of chrge d. (b) The totl electric field t P is long the is. The perpendiculr component of the field t P due to segment 1 is cnceled b the perpendiculr component due to segment 2. de 1 4 To crr out integrtions such s this one, first epress the chrge element d in terms of the other vribles in the integrl. (In this emple, there is one vrible,, so we mde the chnge d 5 l d.) The integrl must be over sclr untities; therefore, epress the electric field in terms of components, if necessr. (In this emple, the field hs onl n component, so this detil is of no concern.) Then, reduce our epression to n integrl over single vrible (or to multiple integrls, ech over single vrible). In emples tht hve sphericl or clindricl smmetr, the single vrible is rdil coordinte.

18 674 CHAPTER 23 Electric Fields 23.7 cont. Anlze Evlute the prllel component of n electric field contribution from segment of chrge d on the ring: (1) de 5 k d e r cos u5k d 2 e 2 1 cos u 2 From the geometr in Figure 23.16, evlute cos u: (2) cos u 5 r /2 ubstitute Eution (2) into Eution (1): d de 5 k e /2 5 k e d /2 2 All segments of the ring mke the sme contribution to the field t P becuse the re ll euidistnt from this point. Integrte to obtin the totl field t P : E 5 3 (3) E 5 k e k d /2 2 k e /2 Q e /2 3 d Finlize This result shows tht the field is zero t 5 0. Is tht consistent with the smmetr in the problem? Furthermore, notice tht Eution (3) reduces to k e Q / 2 if.., so the ring cts like point chrge for loctions fr w from the ring. WHAT IF? uppose negtive chrge is plced t the center of the ring in Figure nd displced slightl b distnce,, long the is. When the chrge is relesed, wht tpe of motion does it ehibit? Answer In the epression for the field due to ring of chrge, let,,, which results in E 5 k eq 3 Therefore, from Eution 23.8, the force on chrge 2 plced ner the center of the ring is F 52 k eq 3 Becuse this force hs the form of Hooke s lw (E. 15.1), the motion of the negtive chrge is simple hrmonic! Emple 23.8 The Electric Field of Uniforml Chrged Disk A disk of rdius R hs uniform surfce chrge densit s. Clculte the electric field t point P tht lies long the centrl perpendiculr is of the disk nd distnce from the center of the disk (Fig ). R d OLUTION Conceptulize If the disk is considered to be set of concentric rings, we cn use our result from Emple 23.7 which gives the field creted b ring of rdius nd sum the contributions of ll rings mking up the disk. B smmetr, the field t n il point must be long the centrl is. Ctegorize Becuse the disk is continuous, we re evluting the field due to continuous chrge distribution rther thn group of individul chrges. r dr Figure (Emple 23.8) A uniforml chrged disk of rdius R. The electric field t n il point P is directed long the centrl is, perpendiculr to the plne of the disk. P Anlze Find the mount of chrge d on ring of rdius r nd width dr s shown in Figure 23.17: d 5s da 5s12pr dr2 5 2psr dr

19 23.6 Electric Field Lines cont. Use this result in the eution given for E in Emple 23.7 (with replced b r nd Q replced b d) to find the field due to the ring: To obtin the totl field t P, integrte this epression over the limits r 5 0 to r 5 R, noting tht is constnt in this sitution: de 5 k e 12psr dr2 1r /2 2 E 5 k e ps 3 R 0 2r dr 1r /2 R 5 k e ps 3 1r /2 d1r Finlize This result is vlid for ll vlues of. 0. We cn clculte the field close to the disk long the is b ssuming R.. ; therefore, the epression in brckets reduces to unit to give us the ner-field pproimtion E 5 2pk e s5 s 2P 0 5 k e ps c 1r /2 21/2 R d 5 2pk e s c R /2 d where P 0 is the permittivit of free spce. In Chpter 24, we obtin the sme result for the field creted b n infinite plne of chrge with uniform surfce chrge densit Electric Field Lines We hve defined the electric field mthemticll through Eution Let s now eplore mens of visulizing the electric field in pictoril representtion. A convenient w of visulizing electric field ptterns is to drw lines, clled electric field lines nd first introduced b Frd, tht re relted to the electric field in region of spce in the following mnner: The electric field vector E is tngent to the electric field line t ech point. The line hs direction, indicted b n rrowhed, tht is the sme s tht of the electric field vector. The direction of the line is tht of the force on positive test chrge plced in the field. The number of lines per unit re through surfce perpendiculr to the lines is proportionl to the mgnitude of the electric field in tht region. Therefore, the field lines re close together where the electric field is strong nd fr prt where the field is wek. These properties re illustrted in Figure The densit of field lines through surfce A is greter thn the densit of lines through surfce B. Therefore, the mgnitude of the electric field is lrger on surfce A thn on surfce B. Furthermore, becuse the lines t different loctions point in different directions, the field is nonuniform. Is this reltionship between strength of the electric field nd the densit of field lines consistent with Eution 23.9, the epression we obtined for E using Coulomb s lw? To nswer this uestion, consider n imginr sphericl surfce of rdius r concentric with point chrge. From smmetr, we see tht the mgnitude of the electric field is the sme everwhere on the surfce of the sphere. The number of lines N tht emerge from the chrge is eul to the number tht penetrte the sphericl surfce. Hence, the number of lines per unit re on the sphere is N/4pr 2 (where the surfce re of the sphere is 4pr 2 ). Becuse E is proportionl to the number of lines per unit re, we see tht E vries s 1/r 2 ; this finding is consistent with Eution Representtive electric field lines for the field due to single positive point chrge re shown in Figure (pge 676). This two-dimensionl drwing shows The mgnitude of the field is greter on surfce A thn on surfce B. A Figure Electric field lines penetrting two surfces. B

20 676 CHAPTER 23 Electric Fields Figure The electric field lines for point chrge. Notice tht the figures show onl those field lines tht lie in the plne of the pge. For positive point chrge, the field lines re directed rdill outwrd. For negtive point chrge, the field lines re directed rdill inwrd. Pitfll Prevention 23.2 Electric Field Lines Are Not Pths of Prticles! Electric field lines represent the field t vrious loctions. Ecept in ver specil cses, the do not represent the pth of chrged prticle moving in n electric field. Pitfll Prevention 23.3 Electric Field Lines Are Not Rel Electric field lines re not mteril objects. The re used onl s pictoril representtion to provide ulittive description of the electric field. Onl finite number of lines from ech chrge cn be drwn, which mkes it pper s if the field were untized nd eists onl in certin prts of spce. The field, in fct, is continuous, eisting t ever point. You should void obtining the wrong impression from twodimensionl drwing of field lines used to describe three-dimensionl sitution. The number of field lines leving the positive chrge euls the number terminting t the negtive chrge. Figure The electric field lines for two point chrges of eul mgnitude nd opposite sign (n electric dipole). onl the field lines tht lie in the plne contining the point chrge. The lines re ctull directed rdill outwrd from the chrge in ll directions; therefore, insted of the flt wheel of lines shown, ou should picture n entire sphericl distribution of lines. Becuse positive test chrge plced in this field would be repelled b the positive source chrge, the lines re directed rdill w from the source chrge. The electric field lines representing the field due to single negtive point chrge re directed towrd the chrge (Fig b). In either cse, the lines re long the rdil direction nd etend ll the w to infinit. Notice tht the lines become closer together s the pproch the chrge, indicting tht the strength of the field increses s we move towrd the source chrge. The rules for drwing electric field lines re s follows: The lines must begin on positive chrge nd terminte on negtive chrge. In the cse of n ecess of one tpe of chrge, some lines will begin or end infinitel fr w. The number of lines drwn leving positive chrge or pproching negtive chrge is proportionl to the mgnitude of the chrge. No two field lines cn cross. We choose the number of field lines strting from n object with positive chrge 1 to be C 1 nd the number of lines ending on n object with negtive chrge 2 to be C u 2 u, where C is n rbitrr proportionlit constnt. Once C is chosen, the number of lines is fied. For emple, in two-chrge sstem, if object 1 hs chrge Q 1 nd object 2 hs chrge Q 2, the rtio of number of lines in contct with the chrges is N 2 /N 1 5 uq 2 /Q 1 u. The electric field lines for two point chrges of eul mgnitude but opposite signs (n electric dipole) re shown in Figure Becuse the chrges re of eul mgnitude, the number of lines tht begin t the positive chrge must eul the number tht terminte t the negtive chrge. At points ver ner the chrges, the lines re nerl rdil, s for single isolted chrge. The high densit of lines between the chrges indictes region of strong electric field. Figure shows the electric field lines in the vicinit of two eul positive point chrges. Agin, the lines re nerl rdil t points close to either chrge, nd the sme number of lines emerges from ech chrge becuse the chrges re eul in mgnitude. Becuse there re no negtive chrges vilble, the electric field lines end infinitel fr w. At gret distnces from the chrges, the field is pproimtel eul to tht of single point chrge of mgnitude 2. Finll, in Active Figure 23.22, we sketch the electric field lines ssocited with positive chrge 12 nd negtive chrge 2. In this cse, the number of lines leving 12 is twice the number terminting t 2. Hence, onl hlf the lines tht leve the positive chrge rech the negtive chrge. The remining hlf terminte on negtive chrge we ssume to be t infinit. At distnces much greter thn b

21 23.7 Motion of Chrged Prticle in Uniform Electric Field 677 A C B Figure The electric field lines for two positive point chrges. (The loctions A, B, nd C re discussed in Quick Quiz 23.5.) Two field lines leve 2 for ever one tht termintes on. 2 the chrge seprtion, the electric field lines re euivlent to those of single chrge 1. Quick Quiz 23.5 Rnk the mgnitudes of the electric field t points A, B, nd C shown in Figure (gretest mgnitude first) Motion of Chrged Prticle in Uniform Electric Field When prticle of chrge nd mss m is plced in n electric field E, the electric force eerted on the chrge is E ccording to Eution If tht is the onl force eerted on the prticle, it must be the net force, nd it cuses the prticle to ccelerte ccording to the prticle under net force model. Therefore, e F 5 E 5 m nd the ccelertion of the prticle is 5 E (23.12) m If E is uniform (tht is, constnt in mgnitude nd direction), the electric force on the prticle is constnt nd we cn ppl the prticle under constnt ccelertion model to the motion of the prticle. If the prticle hs positive chrge, its ccelertion is in the direction of the electric field. If the prticle hs negtive chrge, its ccelertion is in the direction opposite the electric field. ACTIVE FIGURE The electric field lines for point chrge +2 nd second point chrge 2. Pitfll Prevention 23.4 Just Another Force Electric forces nd fields m seem bstrct to ou. Once Fe is evluted, however, it cuses prticle to move ccording to our well-estblished models of forces nd motion from Chpters 2 through 6. Keeping this link with the pst in mind should help ou solve problems in this chpter. Emple 23.9 An Accelerting Positive Chrge: Two Models A uniform electric field E is directed long the is between prllel pltes of chrge seprted b distnce d s shown in Figure A positive point chrge of mss m is relesed from rest t point net to the positive plte nd ccelertes to point net to the negtive plte. (A) Find the speed of the prticle t b modeling it s prticle under constnt ccelertion. OLUTION v 0 E v Conceptulize When the positive chrge is plced t, it eperiences n electric force towrd the right in Figure due to the electric field directed towrd the right. Ctegorize Becuse the electric field is uniform, constnt electric force cts on the chrge. Therefore, s suggested in the problem sttement, the point chrge cn be modeled s chrged prticle under constnt ccelertion. Figure (Emple 23.9) A positive point chrge in uniform electric field E undergoes constnt ccelertion in the direction of the field. d continued

22 678 CHAPTER 23 Electric Fields 23.9 cont. Anlze Use Eution 2.17 to epress the velocit of the prticle s function of position: olve for v f nd substitute for the mgnitude of the ccelertion from Eution 23.12: v 2 f 5 v 2 i 1 2( f 2 i ) (d 2 0) 5 2d v f 5 "2d 5 2 E Å m bd 5 2Ed Å m (B) Find the speed of the prticle t b modeling it s nonisolted sstem. OLUTION Ctegorize The problem sttement tells us tht the chrge is nonisolted sstem. Energ is trnsferred to this chrge b work done b the electric force eerted on the chrge. The initil configurtion of the sstem is when the prticle is t, nd the finl configurtion is when it is t. Anlze Write the pproprite reduction of the conservtion of energ eution, Eution 8.2, for the sstem of the chrged prticle: Replce the work nd kinetic energies with vlues pproprite for this sitution: W 5 DK F e D 5 K 2 K 5 1 2mv f v f 5 Å 2F e D m ubstitute for the electric force F e nd the displcement D: v f 5 Å 21E21d2 m 5 Å 2Ed m Finlize The nswer to prt (B) is the sme s tht for prt (A), s we epect. Emple An Accelerted Electron An electron enters the region of uniform electric field s shown in Active Figure 23.24, with v i m/s nd E N/C. The horizontl length of the pltes is, m. (A) Find the ccelertion of the electron while it is in the electric field. The electron undergoes downwrd ccelertion (opposite E), nd its motion is prbolic while it is between the pltes. OLUTION Conceptulize This emple differs from the preceding one becuse the velocit of the chrged prticle is initill perpendiculr to the electric field lines. (In Emple 23.9, the velocit of the chrged prticle is lws prllel to the electric field lines.) As result, the electron in this emple follows curved pth s shown in Active Figure Ctegorize Becuse the electric field is uniform, constnt electric force is eerted on the electron. To find the ccelertion of the electron, we cn model it s prticle under net force. Anlze The direction of the electron s ccelertion is downwrd in Active Figure 23.24, opposite the direction of the electric field lines. v i î (0, 0) E (,) ACTIVE FIGURE (Emple 23.10) An electron is projected horizontll into uniform electric field produced b two chrged pltes. v

23 ummr cont. Combine Newton s second lw with the mgnitude of the electric force given b Eution 23.8 to find the component of the ccelertion of the electron: F 5 m 5 F m ee 52 m e ubstitute numericl vlues: C21200 N/C kg m/s 2 (B) Assuming the electron enters the field t time t 5 0, find the time t which it leves the field. OLUTION Ctegorize Becuse the electric force cts onl in the verticl direction in Active Figure 23.24, the motion of the prticle in the horizontl direction cn be nlzed b modeling it s prticle under constnt velocit. Anlze olve Eution 2.7 for the time t which the electron rrives t the right edges of the pltes: f 5 i 1 v t t 5 f 2 i v ubstitute numericl vlues: t 5, m 5 v m/s s (C) Assuming the verticl position of the electron s it enters the field is i 5 0, wht is its verticl position when it leves the field? OLUTION Ctegorize Becuse the electric force is constnt in Active Figure 23.24, the motion of the prticle in the verticl direction cn be nlzed b modeling it s prticle under constnt ccelertion. Anlze Use Eution 2.16 to describe the position of the prticle t n time t: f 5 i 1 v i t t 2 ubstitute numericl vlues: f m/s s m cm Finlize If the electron enters just below the negtive plte in Active Figure nd the seprtion between the pltes is less thn the vlue just clculted, the electron will strike the positive plte. We hve neglected the grvittionl force cting on the electron, which represents good pproimtion when deling with tomic prticles. For n electric field of 200 N/C, the rtio of the mgnitude of the electric force ee to the mgnitude of the grvittionl force mg is on the order of for n electron nd on the order of 10 9 for proton. Definitions ummr The electric field E t some point in spce is defined s the electric force F e tht cts on smll positive test chrge plced t tht point divided b the mgnitude 0 of the test chrge: E ; F e 0 (23.7) continued

24 680 CHAPTER 23 Electric Fields Concepts nd Principles Electric chrges hve the following importnt properties: Chrges of opposite sign ttrct one nother, nd chrges of the sme sign repel one nother. The totl chrge in n isolted sstem is conserved. Chrge is untized. Conductors re mterils in which electrons move freel. Insultors re mterils in which electrons do not move freel. Coulomb s lw sttes tht the electric force eerted b point chrge 1 on second point chrge 2 is 12 F 5 k 1 2 e r 2 r^ 12 (23.6) where r is the distnce between the two chrges nd r^ 12 is unit vector directed from 1 towrd 2. The constnt k e, which is clled the Coulomb constnt, hs the vlue k e N? m 2 /C 2. The electric force on chrge plced in n electric field E is Fe 5 E (23.8) At distnce r from point chrge, the electric field due to the chrge is E 5 k e r 2 r^ (23.9) where r^ is unit vector directed from the chrge towrd the point in uestion. The electric field is directed rdill outwrd from positive chrge nd rdill inwrd towrd negtive chrge. The electric field due to group of point chrges cn be obtined b using the superposition principle. Tht is, the totl electric field t some point euls the vector sum of the electric fields of ll the chrges: E 5 k e i i r i 2 r^ i (23.10) The electric field t some point due to continuous chrge distribution is E d 5 k e 3 r 2 r^ (23.11) where d is the chrge on one element of the chrge distribution nd r is the distnce from the element to the point in uestion. Objective Questions denotes nswer vilble in tudent olutions Mnul/tud Guide 1. The mgnitude of the electric force between two protons is N. How fr prt re the? () m (b) m (c) 3.10 m (d) m (e) m 2. Estimte the mgnitude of the electric field due to the proton in hdrogen tom t distnce of m, the epected position of the electron in the tom. () N/C (b) 10 8 N/C (c) N/C (d) 10 6 N/C (e) N/C 3. A ver smll bll hs mss of kg nd chrge of 4.00 mc. Wht mgnitude electric field directed upwrd will blnce the weight of the bll so tht the bll is suspended motionless bove the ground? () N/C (b) N/C (c) N/C (d) N/C (e) N/C 4. An electron with speed of m/s moves into uniform electric field of mgnitude N/C. The field lines re prllel to the electron s velocit nd pointing in the sme direction s the velocit. How fr does the electron trvel before it is brought to rest? () 2.56 cm (b) 5.12 cm (c) 11.2 cm (d) 3.34 m (e) 4.24 m 5. A point chrge of nc is locted t (0, 1.00) m. Wht is the component of the electric field due to the point chrge t (4.00, 22.00) m? () 1.15 N/C (b) N/C (c) 1.44 N/C (d) N/C (e) N/C 6. Two point chrges ttrct ech other with n electric force of mgnitude F. If the chrge on one of the prticles is reduced to one-third its originl vlue nd the distnce between the prticles is doubled, wht is the resulting mgnitude of the electric force between them? () 12F 1 (b) 1 3F (c) 1 6F (d) 3 4F (e) 3 2F 7. Wht hppens when chrged insultor is plced ner n unchrged metllic object? () The repel ech other. (b) The ttrct ech other. (c) The m ttrct or repel ech other, depending on whether the chrge on the insu-